README_EN.md

673. Number of Longest Increasing Subsequence

中文文档

Description

Given an integer array nums, return the number of longest increasing subsequences.

Notice that the sequence has to be strictly increasing.

 

Example 1:

Input: nums = [1,3,5,4,7]
Output: 2
Explanation: The two longest increasing subsequences are [1, 3, 4, 7] and [1, 3, 5, 7].

Example 2:

Input: nums = [2,2,2,2,2]
Output: 5
Explanation: The length of longest continuous increasing subsequence is 1, and there are 5 subsequences' length is 1, so output 5.

 

Constraints:

• 1 <= nums.length <= 2000
• -106 <= nums[i] <= 106

Solutions

similar problem: LIS

cnt array records the number of longest sequence ending in nums[i]

Python3

class Solution:
    def findNumberOfLIS(self, nums: List[int]) -> int:
        maxLen, ans, n = 0, 0, len(nums)
        dp, cnt = [1] * n, [1] * n
        for i in range(n):
            for j in range(i):
                if nums[i] > nums[j]:
                    if dp[j] + 1 > dp[i]:
                        dp[i] = dp[j] + 1
                        cnt[i] = cnt[j]
                    elif dp[j] + 1 == dp[i]:
                        cnt[i] += cnt[j]
            if dp[i] > maxLen:
                maxLen = dp[i]
                ans = cnt[i]
            elif dp[i] == maxLen:
                ans += cnt[i]
        return ans

Java

class Solution {
    public int findNumberOfLIS(int[] nums) {
        int maxLen = 0, ans = 0, n = nums.length;
        int[] dp = new int[n];
        int[] cnt = new int[n];
        for (int i = 0; i < n; i++) {
            dp[i] = 1;
            cnt[i] = 1;
            for (int j = 0; j < i; j++) {
                if (nums[i] > nums[j]) {
                    if (dp[j] + 1 > dp[i]) {
                        dp[i] = dp[j] + 1;
                        cnt[i] = cnt[j];
                    } else if (dp[j] + 1 == dp[i]) {
                        cnt[i] += cnt[j];
                    }
                }
            }
            if (dp[i] > maxLen) {
                maxLen = dp[i];
                ans = cnt[i];
            } else if (dp[i] == maxLen) {
                ans += cnt[i];
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int findNumberOfLIS(vector<int>& nums) {
        int maxLen = 0, ans = 0, n = nums.size();
        vector<int> dp(n, 1), cnt(n, 1);
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < i; ++j) {
                if (nums[i] > nums[j]) {
                    if (dp[j] + 1 > dp[i]) {
                        dp[i] = dp[j] + 1;
                        cnt[i] = cnt[j];
                    } else if (dp[j] + 1 == dp[i]) {
                        cnt[i] += cnt[j];
                    }
                }
            }
            if (dp[i] > maxLen) {
                maxLen = dp[i];
                ans = cnt[i];
            } else if (dp[i] == maxLen) {
                ans += cnt[i];
            }
        }
        return ans;
    }
};

Go

func findNumberOfLIS(nums []int) int {
	maxLen, ans, n := 0, 0, len(nums)
	dp, cnt := make([]int, n), make([]int, n)
	for i := 0; i < n; i++ {
		dp[i] = 1
		cnt[i] = 1
		for j := 0; j < i; j++ {
			if nums[i] > nums[j] {
				if dp[j]+1 > dp[i] {
					dp[i] = dp[j] + 1
					cnt[i] = cnt[j]
				} else if dp[j]+1 == dp[i] {
					cnt[i] += cnt[j]
				}
			}
		}
		if dp[i] > maxLen {
			maxLen = dp[i]
			ans = cnt[i]
		} else if dp[i] == maxLen {
			ans += cnt[i]
		}
	}
	return ans
}

Rust

impl Solution {
    pub fn find_number_of_lis(nums: Vec<i32>) -> i32 {
        let mut max_len = 0;
        let mut ans = 0;
        let n = nums.len();
        let mut dp = vec![1; n];
        let mut cnt = vec![1; n];
        for i in 0..n {
            for j in 0..i {
                if nums[i] > nums[j] {
                    if dp[j] + 1 > dp[i] {
                        dp[i] = dp[j] + 1;
                        cnt[i] = cnt[j];
                    } else if dp[j] + 1 == dp[i] {
                        cnt[i] += cnt[j];
                    }
                }
            }
            if dp[i] > max_len {
                max_len = dp[i];
                ans = cnt[i];
            } else if dp[i] == max_len {
                ans += cnt[i];
            }
        }
        ans
    }
}

...